Two non - integer roots of the equation $(x^{2} + 3 x)^{2} - (x^{2} + 3 x) - 6 = 0$
are

\frac{1}{2} (- 3 + \sqrt{11}), \frac{1}{2} (- 3 - \sqrt{11})

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\frac{1}{2} (- 3 + \sqrt{7}), \frac{1}{2} (- 3 - \sqrt{7})

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\frac{1}{2} (- 3 + \sqrt{21}), \frac{1}{2} (- 3 - \sqrt{21})

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none of these

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Solution

The correct option is C $\frac{1}{2} (- 3 + \sqrt{21}), \frac{1}{2} (- 3 - \sqrt{21})$
Equation is $(x^{2} + 3 x)^{2} - (x^{2} + 3 x) - 6 = 0$
Put $x^{2} + 3 x = y$
Therefore,
$y^{2} - y - 6 = 0$
$(y + 2) (y - 3) = 0$
$y = - 2, 3$
When $y = - 2$
$x^{2} + 3 x = - 2$
$x^{2} + 3 x + 2 = 0$
$(x + 1) (x + 2) = 0$
$x = - 1, - 2$
When, $y = 3$
$x^{2} + 3 x - 3 = 0$
$x = \frac{- 3 \pm \sqrt{9 + 12}}{2}$
$x = \frac{1}{2} (- 3 + \sqrt{21}), \frac{1}{2} (- 3 - \sqrt{21})$

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Two non - integer roots of the equation (x2+3x)2−(x2+3x)−6=0 are

The correct option is C 12(−3+√21),12(−3−√21)Equation is (x2+3x)2−(x2+3x)−6=0Put x2+3x=yTherefore,y2−y−6=0(y+2)(y−3)=0y=−2,3When y=−2x2+3x=−2x2+3x+2=0(x+1)(x+2)=0x=−1,−2When, y=3x2+3x−3=0x=−3±√9+122x=12(−3+√21),12(−3−√21)

Two non - integer roots of the equation $(x^{2} + 3 x)^{2} - (x^{2} + 3 x) - 6 = 0$
are